If $E(X^2)$ is finite then $xP(X>x)$ is eventually decreasing?

Question

Assume $X\ge 0$ be a random variable. Does $E(X^2)<\infty$ implies $f(x):=xP(X>x)$ eventually decreasing?

I proved a bunch of things, however that doesn't lead to anywhere.

$E(X^2)\ge E(X^2\mathbb{I}_{(X>x)})\ge x^2E(\mathbb{I}_{(X>x)})=x^2P(X>x)$ which implies $xP(X>x) \le \frac{E(X^2)}{x}$.

Also we have $\displaystyle \int_0^\infty xP(X>x)dx = \frac12E(X^2)$

Any suggestion/help.

Answer 1

It doesn't have to be decreasing. For example, consider a discrete random variable with $P(X = i) = 2^{-i}$ for $i \in \mathbb{N}_{>0}$ (this is a geometric distribution and all of its moments exist). Then $P(X > x)$ is constant on all the intervals of the form $[n, n + 1)$, so your function $f$ will be increasing within each such interval.